Solve (7p-1)(7p+1)<49p^2

$$( 7 p - 1 ) ( 7 p + 1 ) < 49 p ^ { 2 }$$

Solve for p

$$p\in \mathrm{R}$$

Solution Steps

Consider \left(7p-1\right)\left(7p+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(7p\right)^{2}-1<49p^{2}$$

Expand \left(7p\right)^{2}.

$$7^{2}p^{2}-1<49p^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49p^{2}-1<49p^{2}$$

Subtract 49p^{2} from both sides.

$$49p^{2}-1-49p^{2}<0$$

Combine 49p^{2} and -49p^{2} to get 0.

$$-1<0$$

This is true for any p.

$$p\in \mathrm{R}$$